Sides of a triangle are 6, 10 and x for what value of x is the area of the △…

Key idea: For fixed two sides, the triangle's area is maximized when the angle between those sides is 90°.

• Write the area in terms of the included angle: area = (1/2) * 6 * 10 * sin(theta) = 30 * sin(theta), where theta is the angle between the sides of lengths 6 and 10.

• Since sin(theta) ≤ 1, the area is maximized when sin(theta) = 1, i.e. theta = 90°, so the two given sides are perpendicular.

• When the angle is 90°, the third side x is the hypotenuse: x = sqrt(6^2 + 10^2) = sqrt(136) = 2*sqrt(34) ≈ 11.66.

• Check triangle inequality: 10 - 6 < x < 10 + 6 gives 4 < x < 16, and sqrt(136) ≈ 11.66 lies in this range, so the right triangle is valid.

Conclusion: The area is maximized when x = sqrt(136) = 2*sqrt(34) ≈ 11.66. Because this value does not match any of the numeric choices given, the correct selection is the choice that corresponds to "None of these."

Alternate approach: One can also use Heron's formula to express the area in terms of x and maximize that expression; it leads to the same result.